Question

help! fan and medal!
a ball is thrown vertically upward from the top of a 100-foot tower, with an initial velocity of 20ft/sec. its position function is s(t)=-16t^2+20t+100. what is its velocity in ft/sec when t=1 second?
a. -12
b.-44
c.100
d.-32

what is the average rate of change of y with respect to x over the interval [-2,5] for the function y=3+2?
a. -9
b.3
c.1/3 ( wrong answer )
d.-1

Answer 1

What's stopping you from plugging t = 1 for the first problem?

Answer 2

im getting a choice that is not presented. I don't know what im doing wrong
see:
-16(1)^2+20(1)+100
I get 104

Answer 3

Oh I see, you want the velocity, what you have there is the position, take the derivative and plug t = 1.

Answer 4

what would be the derivative?

Answer 5

Have you done calculus?

Answer 6

I am taking the class. lol I am stuck on this lesson. what would be the answer to my questions ?would you be able to work the problems out and show me how to do it. im confused.

Answer 7

\[s'(t) = -32t-20\] would be your velocity

Answer 8

First derivative of displacement is velocity, the second is acceleration.

Answer 9

where are you he 32 from o.o

Answer 10

Mhm, I just used the power rule

Answer 11

\[\frac{ d }{ dx } x^n = nx^{n-1}\]

Answer 12

more confused lol. ok show me how to solve this. o.o

Answer 13

Ok it seems you haven't got far at all in calculus yet lol

Answer 14

yup

Answer 15

Ok, so can you tell me what you know?

Answer 16

pretty much what I first posted while trying to find an answer to my questions lol.
I have limited time so im fussing over what I am doing wrong lol. o.)

here is my guess:
1- -44
2- -9
am I right?

Answer 17

You can use other methods such as the kinematic equations I suppose, but this question requires you to understand calculus, I gave you the derivative already, so you should be able to figure it out :)

Answer 18

so what I am getting now for my first answer is -32. did I do that right?

Answer 19

No, the derivative is s'(t) = -32t-20, so you still have to plug in t=1 in this equation.

Answer 20

And for your second question average rate of change is \[\frac{ f(b)-f(a) }{ b-a }\]

Answer 21

ok I plugged t=1 in and got -52. not one of my choices.

Answer 22

How did you get that?

Answer 23

\[s(t) = -16t^2+20t+100,~~~s'(t) = -32t+20\] plug t = 1, into s'(t), so find s'(1).

Answer 24

Like I said s(t) is your position, but s'(t) is your velocity

Answer 25

And you're looking for velocity when t = 1

Answer 26

1=-32(1)+20
1= -32+20
1=-12
then what?
t=-12?

Answer 27

Well you leave s'(1) alone, but yes, - 12.

Answer 28

I said -12 earlier lol

Answer 29

You said -32 and -52

Answer 30

But do you understand how that is?

Answer 31

You will need to read your book/ youtube videos I think, then you will really understand it.

Answer 32

yes. thanks. um on the second question a would be -2 and b is 5. right?

Answer 33

Yup

Answer 34

a = -2, b = 5

Answer 35

y=2+3? That's just a horizontal line

Answer 36

is there suppose to be an x there somewhere haha

Answer 37

y=3x+2 im supposed to look for the average rate of y with respect to x over the interval [-2,5] for the function y=3x+2. so this is basically like my slope formula? m=y2-y1/x2-x1?

Answer 38

Yes, similarly

Answer 39

\[\frac{ f(5)-f(-2) }{ 5-(-2) }\]

Answer 40

making my answer for number 2 , --1? lol

Answer 41

Show your work

Answer 42

If it makes more sense, y = 3x+2 just means f(x) = 3x+2

Answer 43

You should not get -1

Answer 44

I got 3

Answer 45

That sounds good!

Answer 46

quicl question. lol
would this answer of mine be correct:
what is the slope for the function y=-5x^2+2 at the point x=1?
a- -5 ( my choice)
b- -10
c--3 ( not the answer)
d-slope no determined

Answer 47

How did you get it

Answer 48

I went with this: y=mx+b formula looked at the slope of my equation which is -5. is my reasoning wrong?

Answer 49

No, all these problems require calculus

Answer 50

You need to use \[\lim_{h \rightarrow 0} \frac{ f(a+h)-f(a) }{ h }\] use this definition to find it

Answer 51

It's not a linear equation it's y = -5x^2+2, if it was y = -5x+2 we could say slope is -5.

Answer 52

how would is et it up though?

Answer 53

\[\lim_{h \rightarrow 0} \frac{ f(1+h)-f(1) }{ h }\]

Answer 54

Can you do the rest?

Answer 55

im assuming h is 0? making my anser =cannot be determined?

Answer 56

No, limit h -> 0 does not mean h = 0, you need to simplify the problem, before you start taking the limit.

Answer 57

Hey, why are they giving you these problems if you haven't learnt it yet..?

Answer 58

I don't know im in a pickle?. lol ok then is it -10?

Answer 59

If you show your work I will tell you whether you're right or not, use the draw tool if you want to, because otherwise to me it seems you're just guessing.

Answer 60

in this question I am lol if h does not equal 0 its just approaching 0? how would I be able to find h before solving o.o

Answer 61

Use your equation you are given, plug it into the "formula" I gave you \[\lim_{h \rightarrow 0} \frac{ -5(1+h)^2+2-(-5(1)^2+2 )}{ h }\]

Answer 62

Now simplify

Answer 63

-3? I put that previously a

Answer 64

Oh, I don't know.

Answer 65

Well since you're out of time I guess I will show you how to do it, the answer I got was -5. @magy33

Answer 66

Note that we keep the limit till we actually take the limit itself.